Now, we have to reason out the values of the two guys on the right, First of all, I have no idea what the values are so I'll use unknowns. Like this:

I examine the chemical equation and see that there is a one-to-one ratio between Ag+ and Cl¯. I know this from the coefficients (both one) of the balanced equation.

That means that the concentrations of the two ions are EQUAL. I can use the same unknown to represent both. Like this:

[Ag+] = x = [Cl¯]

Substituting, we get:

1.77 x 10¯10 = (x) (x)

Now, we take the square root of both sides. I hope I'm not too insulting when I emphasize both sides. I have had lots of people take the square root of the x2 side, but not the other. After the square root, we get:

x = 1.33 x 10¯5 M

This is the answer because there is a one-to-one relationship between the Ag+ dissolved and the AgCl it came from. So, the molar solubility of AgCl is 1.33 x 10¯5 moles per liter.

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

However, there is additional explaining to do when compared to the AgCl example.

Here are the two substances:

Sn(OH)2

Ksp = 5.45 x 10¯27

Ag2CrO4

Ksp = 1.12 x 10¯12

Tin(II) hydroxide

Here is the equation for dissociation:

Sn(OH)2 <:===> Sn2+ + 2 OH¯

and here is the Ksp expression:

Ksp = [Sn2+] [OH¯]2

So far, nothing out of the ordinary. However, that two in front of the hydroxide will come into play real soon.

The ratio between Sn2+ and OH¯ is one-to-two. That means that however much Sn2+ dissolves, we get DOUBLE that amount of OH¯. This is important, so go slow and think it through.

One Sn2+ makes two OH¯. That means that if 'x' Sn2+ dissolves, then '2x' of the OH¯ had to have dissolved.

We now write an equation:

5.45 x 10¯27 = (x) (2x)2

So we have 4x3 = 5.45 x 10¯27

and solving that, we get:

x = 1.11 x 10¯9 M = [Sn]

2 x = 2.22 x 10¯9 M = [OH]

Silver chromate

Here is the usual info:

Ag2CrO4 <===> 2 Ag+ + CrO42¯
Ksp = [Ag+]2 [CrO42¯] = 1.12 x 10¯12

In this problem the cofficient of 2 is on the first ion, not the second. No problem!

1.12 x 10¯12 = (2x)2 (x)

We now have:

4x3 = 1.12 x 10¯12

which is solved to give the answer:

x = 6.54 x 10¯5 M.

Calculate the molar solubility (in mol/L) of a saturated solution of the substance.

The two example substances are:

Bi2S3

Ksp = 1.82 x 10¯99

Cu3(PO4)2

Ksp = 1.93 x 10¯37

Bismuth sulfide

Bi2S3 <===> 2 Bi3+ + 3 S2¯

Ksp = [Bi3+]2 [S2¯]3

1.82 x 10¯99 = (2x)2 (3x)3

108x5 = 1.82 x 10¯99

x = 7.00 x 10¯21 M

Note how 'x' still is the moles of the substance that dissolved.

Copper(II) phosphate

Cu3(PO4)2 <===> 3 Cu2+ + 2 PO43¯

Ksp = [Cu2+]3 [PO43¯]2

1.93 x 10¯37 = (3x)3 (2x)2

108x5 = 1.93 x 10¯37

x = 1.78 x 10¯8 M

PART 3:

Calculating The Effect of a Common Ion on Solubility

Exercise 6: Using the Ksp value from Exercise
1 calculate the solubility of PbSO4 in 0.100 M
Na2SO4.

Let molar solubility of PbSO4 be x.

Ksp = [Pb2+] [SO42-]
= = 1.96´ 10-8

PbSO4(s) + aq
[PbSO4(dissolved)] ®
Pb2+(aq) +
SO42-(aq)

x
x
0.100 +x

Since x is very small compared to 0.100,

0.100 +x is very nearly equal to 0.100.

Ksp = [Pb2+] [SO42-]
= x ´ 0.100 = 1.96´
10-8

\x =
= 1.96´ 10-7
M

Molar solubility of PbSO4 in 0.100 M Na2SO4
= 1.96´ 10-7

Molar solubility of PbSO4 in pure water = 1.40´10-4

Note the solubility is very much reduced in 0.100 M Na2SO4because
of the common ion effect.